Given the vectors \(\mathbf{M}=-10 \mathbf{a}_{x}+4 \mathbf{a}_{y}-8 \mathbf{a}_{z}\) and \(\mathbf{N}=8 \mathbf{a}_{x}+7 \mathbf{a}_{y}-2 \mathbf{a}_{z}\), find: ( \(a\) ) a unit vector in the direction of \(-\mathbf{M}+2 \mathbf{N} ;(b)\) the magnitude of \(5 \mathbf{a}_{x}+\) \(\mathbf{N}-3 \mathbf{M} ;(c)|\mathbf{M}||2 \mathbf{N}|(\mathbf{M}+\mathbf{N})\)

First, let's perform the vector arithmetic to find \(-\mathbf{M}+2 \mathbf{N}\): \begin{align*} -\mathbf{M}+2 \mathbf{N} &= -(-10 \mathbf{a}_{x}+4 \mathbf{a}_{y}-8 \mathbf{a}_{z})+2(8 \mathbf{a}_{x}+7 \mathbf{a}_{y}-2 \mathbf{a}_{z}) \\ &= 10 \mathbf{a}_{x}-4 \mathbf{a}_{y}+8 \mathbf{a}_{z}+16 \mathbf{a}_{x}+14 \mathbf{a}_{y}-4 \mathbf{a}_{z} \\ &= (10+16) \mathbf{a}_{x}+(-4+14) \mathbf{a}_{y}+(8-4) \mathbf{a}_{z} \\ &= 26 \mathbf{a}_{x}+10 \mathbf{a}_{y}+4 \mathbf{a}_{z} \end{align*}The resulting vector is \(26 \mathbf{a}_{x}+10 \mathbf{a}_{y}+4 \mathbf{a}_{z}\).

To find the unit vector, we need to divide the components of the resulting vector by its magnitude. First, find the magnitude: $$|\mathbf{V}|=\sqrt{(26)^2+(10)^2+(4)^2}=\sqrt{26^2 + 10^2 + 4^2} = \sqrt{756}$$Now, find the unit vector:$$\hat{\mathbf{V}} = \frac{26 \mathbf{a}_{x}+10 \mathbf{a}_{y}+4 \mathbf{a}_{z}}{\sqrt{756}}$$

Next, let's perform the vector arithmetic to find \(5 \mathbf{a}_{x}+\mathbf{N}-3\mathbf{M}\): \begin{align*} 5 \mathbf{a}_{x}+\mathbf{N}-3\mathbf{M}&=5 \mathbf{a}_{x}+(8 \mathbf{a}_{x}+7 \mathbf{a}_{y}-2 \mathbf{a}_{z}) -3(-10 \mathbf{a}_{x}+4 \mathbf{a}_{y}-8 \mathbf{a}_{z}) \\ &= 5 \mathbf{a}_{x}+8 \mathbf{a}_{x}+7 \mathbf{a}_{y}-2 \mathbf{a}_{z}+30 \mathbf{a}_{x}-12 \mathbf{a}_{y}+24 \mathbf{a}_{z} \\ &= (5+8+30) \mathbf{a}_{x}+(7-12) \mathbf{a}_{y}+(-2+24) \mathbf{a}_{z} \\ &= 43 \mathbf{a}_{x}-5 \mathbf{a}_{y}+22 \mathbf{a}_{z} \end{align*}The resulting vector is \(43 \mathbf{a}_{x}-5 \mathbf{a}_{y}+22 \mathbf{a}_{z}\).

Find the magnitude of the resulting vector: $$|\mathbf{W}|=\sqrt{(43)^2+(-5)^2+(22)^2}=\sqrt{43^2 + 5^2 + 22^2} = \sqrt{2138}$$

To solve this, first find the magnitude of \(\mathbf{M}\) and \(2\mathbf{N}\): \begin{align*} |\mathbf{M}|=\sqrt{(-10)^2+(4)^2+(-8)^2}=\sqrt{10^2+4^2+8^2}=\sqrt{136} \\ 2\mathbf{N}=2(8 \mathbf{a}_{x}+7 \mathbf{a}_{y}-2 \mathbf{a}_{z})=16 \mathbf{a}_{x}+14 \mathbf{a}_{y}-4 \mathbf{a}_{z}\\ |2\mathbf{N}|=\sqrt{(16)^2+(14)^2+(-4)^2}=\sqrt{16^2+14^2+4^2}=\sqrt{372} \end{align*}Now, find the vector \(\mathbf{M}+\mathbf{N}\): \begin{align*} \mathbf{M}+\mathbf{N}&=(-10 \mathbf{a}_{x}+4 \mathbf{a}_{y}-8\mathbf{a}_{z})+(8 \mathbf{a}_{x}+7 \mathbf{a}_{y}-2 \mathbf{a}_{z}) \\ &= (-10+8) \mathbf{a}_{x}+(4+7) \mathbf{a}_{y}+(-8-2) \mathbf{a}_{z} \\ &= -2 \mathbf{a}_{x}+11 \mathbf{a}_{y}-10 \mathbf{a}_{z} \end{align*}Finally, calculate \(|\mathbf{M}||2 \mathbf{N}|(\mathbf{M}+\mathbf{N})\):$$\sqrt{136} \cdot \sqrt{372}(-2 \mathbf{a}_{x}+11 \mathbf{a}_{y}-10 \mathbf{a}_{z})=-272 \mathbf{a}_{x}+1496 \mathbf{a}_{y}-1360 \mathbf{a}_{z}$$